Question

Prove that \(\lim _{n \rightarrow \infty} \frac{x^{n}}{n !}=0\) for any real \(x\).

Short answer

The proof shows that the limit \( \lim _{n \rightarrow \infty} \frac{x^{n}}{n !} \) is 0 for any real \( x \), due to the divergence between exponential growth and factorial growth, and backed by the Squeeze theorem. Therefore, the base \(x\) cannot grow faster than its factorial denominator, causing the value of the sequence to approach zero.

Step-by-step solution

Assume \(x\) is non-zero, and apply the limit laws

Applying the limit laws, the given expression can be expressed as a sequence: \(\frac{x}{1}, \frac{x^2}{2}, \frac{x^3}{6}, \ldots, \frac{x^n}{n!}\) where \(x\) is non-zero.

Analyzing the elements in the sequence

From the above sequence, we can see that when \(n > |x|\), the ratio between term \(a_n\) and term \(a_{n+1}\) is \( \frac{|x|}{n+1} \) which is less than 1. This indicates the terms in the sequence are getting smaller when \(n > |x|\).

Use the Squeeze theorem

For any \(n > |x|\), all subsequent terms of the sequence are less than \( \frac{|x|}{n+1} \). As \( n \rightarrow \infty \), \( \frac{|x|}{n+1} \rightarrow 0 \). According to the Squeeze theorem, if the terms in the sequence are getting smaller and approach 0, the series converges to 0.

Addressing the case when \(x\) is zero

When \( x = 0 \), the sequence becomes \( \frac{0^{n}}{n !}=0 \) for all \(n\). Thus the limit as \(n \rightarrow \infty\) is also 0.